What is the integral of 2/(x^3sqrt(x^4-1)) ?

The integral of 2/(x^3sqrt(x^4-1)) is (sqrt(x^4-1))/(x^2)+C

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integral of 2/(x^3sqrt(x^4-1))

\int \frac{2}{x ^{3} x ^{4} - 1} d x

\frac{x ^{4} - 1}{x ^{2}} + C

Solution steps

\int \frac{2}{x ^{3} x ^{4} - 1} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{1}{x ^{3} x ^{4} - 1} d x

Apply u-substitution

: \int \frac{1}{2 u ^{2} u ^{2} - 1} d u

= 2 \cdot \int \frac{1}{2 u ^{2} u ^{2} - 1} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{u ^{2} u ^{2} - 1} d u

Apply Trigonometric Substitution

: \int cos (v) d v

= 2 \cdot \frac{1}{2} \cdot \int cos (v) d v

Use the common integral:

\int cos (v) d v = sin (v)

= 2 \cdot \frac{1}{2} sin (v)

Substitute back

= 2 \cdot \frac{1}{2} sin (arcsec (x^{2}))

Simplify

2 \cdot \frac{1}{2} sin (arcsec (x^{2})) : \frac{x ^{4} - 1}{x ^{2}}

= \frac{x ^{4} - 1}{x ^{2}}

Add a constant to the solution

= \frac{x ^{4} - 1}{x ^{2}} + C

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What is the integral of 2/(x^3sqrt(x^4-1)) ?
The integral of 2/(x^3sqrt(x^4-1)) is (sqrt(x^4-1))/(x^2)+C